Can generators run in-place?

[popizdeh]

What would happen if both Input<Buffer> and Output<Buffer were pointing to the same memory? I have a feeling this is not safe...

[peteraarts811]

Currently we actually do this.
However, you need to take this situation into account when creating your schedule.
Currently our solution is somewhat like the following:

1. intermediate_buffers = some_function1(input_buffers)
2. output_buffers = some_function2(intermediate_buffers )

In our schedule we do a compute_root on intermediate_buffers,
after which it is safe to write the output_buffers (which will actually overwrite the input_buffers).

You do need to take into account though that the auto-scheduler will probably not help you (as far as I know..).

[popizdeh]

Thanks @peteraarts811 but is this truly in-place? Doesn't compute_root() on intermediate buffers allocate another image sized buffer?

[steven-johnson]

Depends on how you define 'in-place': if that means "the input buffer and output buffer can point to the same memory", then yes. If you also mean "there is no memory allocation done for any intermediate computation", then no.

It's worth pointing out that achieving the latter is challenging regardless of the coding language you use; one can certainly write trivial examples that can get by with only register usage, but that's unlikely to be practical for a huge swath of algorithms.

[popizdeh]

Thanks for the clarification @steven-johnson, I was hoping to get the latter but was very much expecting to get the answer you provided.

To give you a bit more information I'm dealing with edge blur where edge pixels are just a fraction of the whole image, so reading and writing each pixel guarantees poor performance regardless of how you schedule it. But I understand this type of algorithm is not the common case for Halide.